There is no solution ,<span>a+c=-10;b-c=15;a-2b+c=-5 </span>No solution System of Linear Equations entered : [1] 2a+c=-10
[2] b-c=15
[3] a-2b+c=-5
Equations Simplified or Rearranged :<span><span> [1] 2a + c = -10
</span><span> [2] - c + b = 15
</span><span> [3] a + c - 2b = -5
</span></span>Solve by Substitution :
// Solve equation [3] for the variable c
<span> [3] c = -a + 2b - 5
</span>
// Plug this in for variable c in equation [1]
<span><span> [1] 2a + (-a +2?-5) = -10
</span><span> [1] a = -5
</span></span>
// Plug this in for variable c in equation [2]
<span><span> [2] - (-? +2b-5) + b = 15
</span><span> [2] - b = 10
</span></span>
// Solve equation [2] for the variable ?
<span> [2] ? = b + 10
</span>
// Plug this in for variable ? in equation [1]
<span><span> [1] (? +10) = -5
</span><span> [1] 0 = -15 => NO solution
</span></span><span>No solution</span>
Using a discrete probability distribution, it is found that:
a) There is a 0.3 = 30% probability that he will mow exactly 2 lawns on a randomly selected day.
b) There is a 0.8 = 80% probability that he will mow at least 1 lawn on a randomly selected day.
c) The expected value is of 1.3 lawns mowed on a randomly selected day.
<h3>What is the discrete probability distribution?</h3>
Researching the problem on the internet, it is found that the distribution for the number of lawns mowed on a randomly selected dayis given by:
Item a:
P(X = 2) = 0.3, hence, there is a 0.3 = 30% probability that he will mow exactly 2 lawns on a randomly selected day.
Item b:

There is a 0.8 = 80% probability that he will mow at least 1 lawn on a randomly selected day.
Item c:
The expected value of a discrete distribution is given by the <u>sum of each value multiplied by it's respective probability</u>, hence:
E(X) = 0(0.2) + 1(0.4) + 2(0.3) + 3(0.1) = 1.3.
The expected value is of 1.3 lawns mowed on a randomly selected day.
More can be learned about discrete probability distributions at brainly.com/question/24855677